Problem: Divide the following complex numbers. $ \dfrac{-9-7i}{3-i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${3+i}$ $ \dfrac{-9-7i}{3-i} = \dfrac{-9-7i}{3-i} \cdot \dfrac{{3+i}}{{3+i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-9-7i) \cdot (3+i)} {(3-i) \cdot (3+i)} = \dfrac{(-9-7i) \cdot (3+i)} {3^2 - (-1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-9-7i) \cdot (3+i)} {(3)^2 - (-1i)^2} = $ $ \dfrac{(-9-7i) \cdot (3+i)} {9 + 1} = $ $ \dfrac{(-9-7i) \cdot (3+i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-9-7i}) \cdot ({3+i})} {10} = $ $ \dfrac{{-9} \cdot {3} + {-7} \cdot {3 i} + {-9} \cdot {1 i} + {-7} \cdot {1 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{-27 - 21i - 9i - 7 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{-27 - 21i - 9i + 7} {10} = \dfrac{-20 - 30i} {10} = -2-3i $